Is MIT Prof. Lewin wrong about Kirchhoff's law?

[vanhees71]

It may well be that there are powerfull analyses techniques in EE to treat such problems for practical purposes. Physics is not about such techniques but about fundamental properties of nature, and there are Maxwell's equations for the fields and the Lorentz-force Law for the forces of em. fields on charged particles (let alone the self-force problem at this point, which makes things more complicated). I think the here considered experiments can be well explained and understood using only the very basic principles, i.e., Maxwell's equations. One doesn't need more advanced calculational tools of EE, which do not add to the understanding of the basic principles but are practical tools for more complicated problems. They are for sure not part of an introductory (or even advanced) EM lecture for physicists but for more specialized engineers.

 

[yungman]

It may well be that there are powerfull analyses techniques in EE to treat such problems for practical purposes. Physics is not about such techniques but about fundamental properties of nature, and there are Maxwell's equations for the fields and the Lorentz-force Law for the forces of em. fields on charged particles (let alone the self-force problem at this point, which makes things more complicated). I think the here considered experiments can be well explained and understood using only the very basic principles, i.e., Maxwell's equations. One doesn't need more advanced calculational tools of EE, which do not add to the understanding of the basic principles but are practical tools for more complicated problems. They are for sure not part of an introductory (or even advanced) EM lecture for physicists but for more specialized engineers.

It is not, the physical property of what I described about capacitor inductors is absolutely based on EM wave behavior in the transmission line media. The analogy is light wave in optics, I had some very limited knowledge, something like avoiding reflection using coating of quater wave thickness and make the medium thickness into half wavelength and put into another medium and it literally disappeared and that wave length because of the transformation.

It is not a special technique. the voltage induced into the loop of Levine's experiment is REAL! Physics is about dealing with real things and behavior, you cannot just disregard and chuck into " non conservative". This seems to be a big gapping hole in electrodynamics compare to electromagnetics studied in EE.

This happened to be a very easy circuit. I am going to draw up a multi pole filter used in microwave design only make up of sections of copper lines on the pcb. If Levine look at it, it is only a wire from point A to point B...or better yet, just a point C...a note. But this circuit PHYSICALLY IS a real filter that as effective as the real circuit components...In fact even more because the real physical cap is not a cap in RF! It is physically not a cap at RF! BECAUSE Physically, there is parasitic components inside the cap, the conductance of the dielect, inductance of a straight wire, skin effect due to EM attenuation in a conductor...Every bit of it is Electrodynamics and every bit of it is physics.

The most power tool in RF design is using the Smith Chart, which is based on EM wave behavior when the wave encounter a boundary and have reflection trabelling backwards. It is all electromagnetics, not a technique. They should really put this part into the physics electrodynamic class for the graduate level.

 

[vanhees71]

What is wrong with explaining the issue with Farday's Law (one of Maxwell's equations)? What was, in your opinion, wrong with this very simple explanation, I've given this morning?

I have not considered any capacitors, but that's also not a big deal with the there explaned technique of directly applying simply Maxwell's equations to simple circuits. So what's the problem to explain simple things in the most simple way and not simpler?

 

[Studiot]

I wonder what post#1000 will say in this thread?

Perhaps Danger will have a view.

 

[Andrew Mason]

Prof. Lewin mentions Lenz's Law but does he take it into account in his analysis of the circuit? I don't see where he does. Or perhaps I am missing something. He seems to be treating it as negligible and I think that is a big mistake.

Let's consider the situation where Lenz' Law is not a factor: no induced current. Suppose a very small switch is inserted at the bottom of the circuit and opened. The voltmeter on the left measures the potential from the left side of the open switch to the middle top of the circuit - the 100 ohm side - and the voltmeter on the right measures the potential from the middle top to the right side of the switch - the 900 ohm side. What would the voltmeters read when the solenoid is powered up?

Applying Faraday's law, the changing flux from that large solenoid induces an emf around the path of the resistance and wires + the open switch such that \oint E\cdot dl = - d\phi/dt. What would the voltmeters read? I would think the left one would read the same as the right one because the paths are the same length and the E field would be symmetrical between the sides. I don't see why the value of the resistance would affect the induced voltage since there is no current (switch open). The potential depends on the path length, not the value of the resistance.

Now suppose that the switch is closed and the experiment repeated. As soon as the magnetic field of the solenoid starts building up, an emf is induced in the circuit such that emf = \oint E\cdot dl = -d\phi/dt. Now the induced emf starts current flowing in the circuit. Suppose at time t very shortly after the solenoid current begins that total induced emf is 1 volt. The rate of change of that current is quite high at the beginning so there is a high back-emf (Lenz' law) opposing that increase in current so, while the induced voltage from the solenoid field at time t is 1 volt, the current is not 1/1000 A. = 1 mA. It is much less. The two voltages measured on the voltmeters cannot add up to 1 volt.

AM

 

[yungman]

Levine's circuit is as simple as it gets. Try this one.

I design this kind of circuits all the time. Top is the circuit in inductors and capacitors. It is a 6 pole low pass filter that can be design say to cut at 2 GHz. At 4GHz, it can be design to attenuate the signal to 1/20.

The bottom part is what I design the same filter with just the copper clad pattern on a circuit boards. Look just like a pattern! It is every bit as real as the circuit on top. My guess when Levine look at it, he'll be saying it's only a wire connected from A to B. Or better yet, it is a note "C"!

This design is not just an application. It is design according to electromagnetic of guided structure with forward and backward EM wave. Design is making use of boundary condition at B and use the mismatch reflection to manipulate the impedance at each point of the line. Everything is Electromagnetic in the books!

So these all chuck to "non conservative" again? Path dependent again?

Back to Levine, The induced emf is every bit a REAL voltage source. Try to convince someone that put the fingers across the terminals and wet his/her pants if you induce a high enough voltage into it! Ask whether that is real or not! How do physicist justify this, non conservative?

the example I gave is just a very simple circuit that is made of pure passive components. try some with active components and you'll see. So most of the advanced electronics is just non conservative?


I understand Lavine taught a beginer's class, but when he involked challenging professors in EE, he cross the line. He must have grapped the professor in DSP, AI ( artificial intel). those that don't know $#%t about RF design. matter of fact, a lot don't even know analog design!


You have to take into account of all the components, physical and equivlent! Or else, you really get stuck.

 

[stevenb]

Prof. Lewin mentions Lenz's Law but does he take it into account in his analysis of the circuit? I don't see where he does. Or perhaps I am missing something. He seems to be treating it as negligible and I think that is a big mistake.

Why would you say this without first doing an order of magnitude estimate to see if you are right? Assuming I'm interpreting your concern correctly, I think if you run the numbers, you'll see that it is not a big mistake to neglect this effect. There are probably about a 6 orders of magnitude buffer here.

For example, in the experiment I did and documented well, I estimate a loop self-inductance to be of the order of 0.1 microHenries. With this inductance value and a 1 V loop emf, the current change will initially be 10 megaAmps per second. So, how long does it take to get to 1 mA? It's on the order of 100 picoseconds. Now consider the circuit time constant (L/R) which is of the order of 100 picoseconds. So round this up to a nanosecond (1 GHz frequencies) and consider that the MIT old lab scopes are probably 10 MHz or maybe 100 MHz scopes. Will the scope even pick this up? And, even if it did, who cares since it's practically instantaneous for our purposes? The interesting effects are happening on the order of milliseconds, as clearly mentioned by Lewin in the video. With 6 orders of magnitude faster response time, we should be safe neglecting the effect.

 

[cabraham]

Can you show me a case of two parallel circuits do not have the same voltage across it. Please put in the equvalent voltage source also.

The transformer is such a case. A power source, ac constant voltage amplitude & frequency, sine wave, excites the primary of a xfmr. The secondary is connected to a heater load. The secondary output is 120V rms, & 10A rms. The secondary winding resistance is 0.1 ohm. At a full load of 10 amp, the voltage drop in the sec winding is Iload*Rsecwdg = 10A*0.1 ohm = 1.0V.

So if we place a voltmeter across the sec terminals, we read 119V rms. But what is the voltage across the same 2 points if the path is the copper secondary winding. Answer: 1.0V. So if we measure the sec terminal voltage outside the sec winding it is 119V rms, but inside the copper sec it is 1.0V rms.

Of course when analyzing in terms of equiv circuit, we can add the source. THe source is 120V rms, & 1.0V rms is dropped across the sec winding resistance, leaving 119V rms, which is what a VM connected across the sec terminals displays. KVL is aok when the source is added to the equiv circuit. I've said that 100 times.

But the voltage from a to b along the copper path is still 1.0V. By definition, J = sigma*E, & Vab = integral E*dl. The E field inside the conductor is very small, so that the line integral is also small, resulting in a small voltage. The 120V rms source is not present inside the copper. The sec winding insulation, however, is subjected to the full 120V stress divided by the number of turns.

Outside the sec copper winding, the 120V source is present, but not inside. Two parallel paths have differing potentials. Or look at it another way.

How is voltage defined? Is it not the work per unit charge transporting said charge from a to b? It is , of course. IF the external load resistor across the sec in 11.9 ohm, then transporting 10A (10 coulomb/second) through 11.9 ohms requires 119 joule/coulomb, which is 120V. But transporting the same 10A (the sec winding is in series w/ the load) through the 0.1 ohm sec winding requires 1.0 joule/coulomb = 1.0V rms.

It requires much more work per unit charge to transport said charge from a to b when the path is outside the sec copper winding, as opposed to inside. So here is a prime example of 2 paths in parallel w/ differing voltages.

Again, Dr. Lewin's style of teaching is not among my favorite. His methods of explaining things differ from mine. Hopefully you can understand what I've been explaining. BR.

Claude

 

[Studiot]

+3 Cabraham, very well put.

Which brings us nicely back to the point Maxwell clearly understood and made himself viz

The whole beauty of KVL, properly formulated, is that it allows us to completely sidestep the issue of what goes on inside a transformer, battery or indeed any source of EMF.

 

[yungman]

The transformer is such a case. A power source, ac constant voltage amplitude & frequency, sine wave, excites the primary of a xfmr. The secondary is connected to a heater load. The secondary output is 120V rms, & 10A rms. The secondary winding resistance is 0.1 ohm. At a full load of 10 amp, the voltage drop in the sec winding is Iload*Rsecwdg = 10A*0.1 ohm = 1.0V.

So if we place a voltmeter across the sec terminals, we read 119V rms. But what is the voltage across the same 2 points if the path is the copper secondary winding. Answer: 1.0V. So if we measure the sec terminal voltage outside the sec winding it is 119V rms, but inside the copper sec it is 1.0V rms.

Of course when analyzing in terms of equiv circuit, we can add the source. THe source is 120V rms, & 1.0V rms is dropped across the sec winding resistance, leaving 119V rms, which is what a VM connected across the sec terminals displays. KVL is aok when the source is added to the equiv circuit. I've said that 100 times.
That is not true, emf sources are distributed throughout the winding, say you have 120 turns in the secondary, each turn is only going to give you (1 - 1/120)V. Point is the voltage cannot be separated from the winding. You cannot measure as if 1V drop across the coil and 120 equivalent generator. That is the whole thing about distribution.
But the voltage from a to b along the copper path is still 1.0V. By definition, J = sigma*E, & Vab = integral E*dl. The E field inside the conductor is very small, so that the line integral is also small, resulting in a small voltage. The 120V rms source is not present inside the copper. The sec winding insulation, however, is subjected to the full 120V stress divided by the number of turns.

Outside the sec copper winding, the 120V source is present, but not inside. Two parallel paths have differing potentials. Or look at it another way.
There is no two parallel path, it is one path with infinite micro voltage source per dl.
How is voltage defined? Is it not the work per unit charge transporting said charge from a to b? It is , of course. IF the external load resistor across the sec in 11.9 ohm, then transporting 10A (10 coulomb/second) through 11.9 ohms requires 119 joule/coulomb, which is 120V. But transporting the same 10A (the sec winding is in series w/ the load) through the 0.1 ohm sec winding requires 1.0 joule/coulomb = 1.0V rms.

It requires much more work per unit charge to transport said charge from a to b when the path is outside the sec copper winding, as opposed to inside. So here is a prime example of 2 paths in parallel w/ differing voltages.

Again, Dr. Lewin's style of teaching is not among my favorite. His methods of explaining things differ from mine. Hopefully you can understand what I've been explaining. BR.

Claude

......

 

[yungman]

+3 Cabraham, very well put.

Which brings us nicely back to the point Maxwell clearly understood and made himself viz

The whole beauty of KVL, properly formulated, is that it allows us to completely sidestep the issue of what goes on inside a transformer, battery or indeed any source of EMF.

No you cannot. Show me how.

 

[yungman]

Guys, it is becoming clearer and clearer the difference between my issue with this whole thread vs a lot of you guys in the electrodynamic side is the existing of the voltage source or emf source. Please feel free to disagree with me and adress this directly, you don't have to be diplomatic. If I am wrong, I'll learn. I just don't see how you can ignor the existence of the induced voltage source and go through hoops to justify the non existence of a REAL voltage source.

Yes Maxwell's eq. have no provision for induced voltage source. BUT Maxwell's equations to a big extend are formed by observation. Just like in time varying case of \nabla \times \vec H where the \frac {\partial \vec D}{\partial t} needed to be added because \nabla \cdot \vec J = \frac {\partial \vec \rho_v}{\partial t}.

The secondary voltage of the transformer case by Cabraham is even more obvious. With multiple turn secondary, you cannot even measure the two paths and show different voltages like in Levine's experiment. The 1V drop is a theractical voltage, and cannot be measure by any known means because the voltage sources are distributed throughout the entire secondary. Levine uses a single turn transformer, at least you can play some games to measure the voltage of the two separate paths like what I did because the probe ground is a single turn also.

All Electromagnetics textbooks use equivalent voltage and components. How can electrodynamic accomadate all the electronic theories which is a big part of RF and analog world and we actually have to have two branches of EM!

 

[cabraham]

From yungman: "That is not true, emf sources are distributed throughout the winding, say you have 120 turns in the secondary, each turn is only going to give you (1 - 1/120)V. Point is the voltage cannot be separated from the winding. You cannot measure as if 1V drop across the coil and 120 equivalent generator. That is the whole thing about distribution."

I don't think so. How can there be distributed voltage sources inside the copper? Ohm's law must be met, i.e. J = sigma*E. Also, Faraday's Law holds inside as well as outside the wire. FL states that curl E = -dB/dt. So the E field in the wire which gives rise to a "distributed source of emf", must have curl to do so. In order to have a curly E field inside the wire, there must be a "dB/dt" inside the wire as well. For a perfect superconductor, B does not enter the wire. For an ordinary conductor, there is a small B field entering the wire.

The reason there is very little emf source inside the copper is because there is very little B field inside the CU, & hence very little rotational (curly) E field, & hence very little emf "source".

I find it most helpful to look at this in terms of Lorentz force. The xfmr core B field is accompanied by an E field, both sinusoidal. The E & B fields exert a force on the free electrons in the secondary Cu winding & the resistive heater which is the external load. There is 120V rms of emf acting on the circuit loop. But a volt is just a joule/coulomb, so there is 120 joules of energy expended transporting one coulomb of charge around the loop one time. OK?

The Cu sec winding & the heater have resistance. Energy is converted to heat per P = I^2*R. The 10A through the sec Cu, Rsec=0.1 ohm, results in 1.0 joules lost for every coulomb. The 10A through the 11.9 ohm heater results in 119 joules per coulomb. Thus the 120V loop emf is divided as follows. The voltage across the terminals a & b along the Cu path is 1.0V, & along the external heater path is 119V.

When drawing an equivalent circuit for computational purposes, it is perfectly fine to use a lumped parameter representation. A 120V source in series with the 0.1 ohm Cu sec resistance in series w/ the 11.9 ohm heater load serve to facilitate analysis. The right answer is obtained w/ this approach.

An equivalent circuit is by definition, mathematically precise enough for computational purposes. It is not always a 1 to 1 replica of the true physical phenomena. I cannot accept the "distributed source of emf" because there is little B field inside the Cu. The distributed sources of emf, infinitessimal in length have to exist on the outside of the Cu.

If I've erred feel free to show me where. But what I've stated has been known through observation & reaffirmation for over a century. To claim otherwise means that over a century of the best scientific minds using instruments have missed something a modern critic has perceived intuitively. I don't think so. The case for distributed emf sources inside the Cu is pretty weak, actually more like non-existant.

Claude

 

[yungman]

From yungman: "That is not true, emf sources are distributed throughout the winding, say you have 120 turns in the secondary, each turn is only going to give you (1 - 1/120)V. Point is the voltage cannot be separated from the winding. You cannot measure as if 1V drop across the coil and 120 equivalent generator. That is the whole thing about distribution."

I don't think so. How can there be distributed voltage sources inside the copper? Ohm's law must be met, i.e. J = sigma*E. Also, Faraday's Law holds inside as well as outside the wire. FL states that curl E = -dB/dt. So the E field in the wire which gives rise to a "distributed source of emf", must have curl to do so. In order to have a curly E field inside the wire, there must be a "dB/dt" inside the wire as well. For a perfect superconductor, B does not enter the wire. For an ordinary conductor, there is a small B field entering the wire.

The reason there is very little emf source inside the copper is because there is very little B field inside the CU, & hence very little rotational (curly) E field, & hence very little emf "source".

I find it most helpful to look at this in terms of Lorentz force. The xfmr core B field is accompanied by an E field, both sinusoidal. The E & B fields exert a force on the free electrons in the secondary Cu winding & the resistive heater which is the external load. There is 120V rms of emf acting on the circuit loop. But a volt is just a joule/coulomb, so there is 120 joules of energy expended transporting one coulomb of charge around the loop one time. OK?

The Cu sec winding & the heater have resistance. Energy is converted to heat per P = I^2*R. The 10A through the sec Cu, Rsec=0.1 ohm, results in 1.0 joules lost for every coulomb. The 10A through the 11.9 ohm heater results in 119 joules per coulomb. Thus the 120V loop emf is divided as follows. The voltage across the terminals a & b along the Cu path is 1.0V, & along the external heater path is 119V.

When drawing an equivalent circuit for computational purposes, it is perfectly fine to use a lumped parameter representation. A 120V source in series with the 0.1 ohm Cu sec resistance in series w/ the 11.9 ohm heater load serve to facilitate analysis. The right answer is obtained w/ this approach.

An equivalent circuit is by definition, mathematically precise enough for computational purposes. It is not always a 1 to 1 replica of the true physical phenomena. I cannot accept the "distributed source of emf" because there is little B field inside the Cu. The distributed sources of emf, infinitessimal in length have to exist on the outside of the Cu.

If I've erred feel free to show me where. But what I've stated has been known through observation & reaffirmation for over a century. To claim otherwise means that over a century of the best scientific minds using instruments have missed something a modern critic has perceived intuitively. I don't think so. The case for distributed emf sources inside the Cu is pretty weak, actually more like non-existant.

Claude

Thanks for taking the time. I see your rationel on the voltage source cannot be inside because curl E is small inside the copper wire if \frac{\partial \vec B}{\partial t}is small in good conductor.

But where is the voltage come from? Len's law is not about current only, it is a physical voltage because of all the formulas on voltage ratio regardless of load. That is my whole thing about this whole thread. Where is the voltage come from? It is sure not at the end of the wire between the heater load and the secondary winding. The fact that you can physically meansure continuous incrememtal increase of voltage along the secondary winding, that has to show some voltage source along the wire. How does Maxwell's equation justify this voltage?

I think this is the bottom line my disaggrement with Levine and some of the people here. How do you account for the voltage? Took me a long time to even realize you guys don't do voltage! No wonder we are talking in different language! Levine only has one turn secondary that make measurement along the loop difficult as proofen from my experiment. The example you gave about the secondary of the transformer make it very very easy to physically measure the voltage along the winding. I actually have a Marklin train set that has a transformer supply and the speed is adjusted by physically sweeping on the surface of the secondary winding. That is also how the variac work.

It's is so clear that there is a voltage source in Levine's case and your transformer case and there is no way to take it out or ignor it. I truly don't care about the definition of KVL or conservative and all. It is the voltage! It would be nice it you can explain this to me.

Thanks

Alan

 

[Antiphon]

Folks, I'm trying my level best to stay out of this thread but like Michael Corleone in Godfather 3, I feel I am being pulled back in.

Let's clarify something right from the start. There is electromagnetics which has E, H and J (current density). This is physics.

Then there is a whole 'nother field of study called circuit analysis. In this discipline there are no electric and magnetic fields. There is electric current I and voltage V.

In this discipline KVL has been defined as the sum of potentials around a loop adding to zero. This is an axiomatic definition of KVL *in circuit theory*. It is NOT physics.

*Whenever* a field of any kind is discussed in relation to a circuit, it is a physics problem, not a problem in circuit theory.

EE professors do things that physicists do not do in order to arrive at self-consistent useful theories like circuit theory that are NOT physics.

Every day EEs design power supplies using ideal transformers shunted by inductors. No physicist would ever think to connect a battery to an ideal transformer and get stepped up DC to come out the secondary. But this is what a transformer *is* in circuit analysis. A EE shunts this DC-capable transformer with an inductor to arrive at a description of what a more realistic transformer does. This isn't even close to the correct physics but it's what you do to design a transformer.

When physicists in this forum talk about the obvious flaw of leaving out induction from the KVL, it simply means they aren't familiar with modern circuit theory as practiced by the circuit professionals, Electrical Engineers.

Now, professor Lewin: he invoked the KVL of circuit theory then switched over to physics to make a point with the students. His physics and teaching style are wonderful. It's just that there is an Electrical Engineering KVL which has no fields and no induction, and there is physics with induction and potententials.

That's all.

 

[yungman]

Hey Antiphon, don't stay out of this thread. Don't look at it as confrontation, rather as an exchange of opinion. Statement might get a little hot...well passionate. But hey, that's the reason we are here at the first place spending hours...weeks in my case on this thread!



Anyway, get back to the subject:

I thought PHYSICS is about explaining the physical phenomena. For any laws and theory like Maxwell's, Continuity etc. It is has work in all cases. You cannot exclude certain situation. I don't believe there should be electrodynamics for physicist and electromagnetic for EE. In my book, electronics is a branch of physic...isn't it? Everything is EM in electronics except pure DC circuits. In order to call any theory and law, it has to apply to all situation UNLESS it SPECIFIES an exception. BUT I don't see exception in Maxwell's equations. In this case, the induced emf is every bit real and physical.

So in order to convince EE like me that a voltage source does not exist, that equivalent circuit do not exist, physicist should come up with a reason for that to explain why. In this case, what is the reason that the physicist refuse to acknowledge the voltage source in Levine's resistors loop and the secondary winding of the transformer.

There are induced voltage in different cases, not just these two examples. More important is the radio EM wave receive by antenna. It actually generate a voltage on the antenna out of the thin air! This is how the antenna receive the EM signal from the air to be demodulated and amplified. This is a big field in the industry, you cannot just ignor this and call this non conservative.

 

[Antiphon]

Yungman, I will make one more post to this thread for your sake but then I will stay out.

Physics is indeed about the physical phenomena. Electrodynamics is not the same discipline as electromagnetics though because the tools and the problems are different. In electrodynamics one might study the field of a uniformly moving electron. In electromagnetics one might employ a sheet of magnetic charges and currents to arrive at the solution to a problem in antenna design. They are both correct theories but one studies the real world while the other is a set of tools that help one solve applied problems.

I hate to dissapoint you but circuit theory is distinct from Maxwell's equations and predicts incorrect results as soon as it's assumptions are violated. One of those assumptions (induction taking place outside the terminals of an inductor) was violated by professor Lewin to make his point.

Engineering is about building and using mathematical approximations to physics so as to be practical. Physics is finding out how nature actually works. Not the same thing.

 

[yungman]

Yungman, I will make one more post to this thread for your sake but then I will stay out.

Physics is indeed about the physical phenomena. Electrodynamics is not the same discipline as electromagnetics though because the tools and the problems are different. In electrodynamics one might study the field of a uniformly moving electron. In electromagnetics one might employ a sheet of magnetic charges and currents to arrive at the solution to a problem in antenna design. They are both correct theories but one studies the real world while the other is a set of tools that help one solve applied problems.

I hate to dissapoint you but circuit theory is distinct from Maxwell's equations and predicts incorrect results as soon as it's assumptions are violated. One of those assumptions (induction taking place outside the terminals of an inductor) was violated by professor Lewin to make his point.

Engineering is about building and using mathematical approximations to physics so as to be practical. Physics is finding out how nature actually works. Not the same thing.

Well try to hang around. You are one of the very few here that can see things from both side.

It just take me so long to realize physicist missing this whole big chunk. It really never dawn on me to even question that, no wonder all the drawing, all the talk about voltage source is like dropping into deep well! Lewin was so obviously wrong and very few here even pick that up. It is you that start talking about physical size of circuit and difference between circuit theory vs electrodynamics that make me stop and start thinking about what is missing in this 300 posts!

But I still think if the physicist is all about talking theory and definition, they need to cover the phenomena in electronics because electronics in only a small subset of physics Maxwell has to agree with circuit theory also. Physist do step into electronics like Levin did. But they really need to learn electronics before they make a fool of themself like Levin. Just like kenetics, heat and other fields, they have a law or a theorem, they better be able to apply to ALL cases.

 

[Antiphon]

Dammit I'm still here and I don't want to be.

The professor did not make a fool of himself. He knew exactly what was going on and only wanted to present a good puzzle for the students to ponder.

Now I really am done here.

 

[yungman]

Dammit I'm still here and I don't want to be.

The professor did not make a fool of himself. He knew exactly what was going on and only wanted to present a good puzzle for the students to ponder.

Now I really am done here.

I am sure there will be a lot of people come into correct me!

 

[hikaru1221]

Let's consider the situation where Lenz' Law is not a factor: no induced current. Suppose a very small switch is inserted at the bottom of the circuit and opened. The voltmeter on the left measures the potential from the left side of the open switch to the middle top of the circuit - the 100 ohm side - and the voltmeter on the right measures the potential from the middle top to the right side of the switch - the 900 ohm side. What would the voltmeters read when the solenoid is powered up?

Applying Faraday's law, the changing flux from that large solenoid induces an emf around the path of the resistance and wires + the open switch such that \oint E\cdot dl = - d\phi/dt. What would the voltmeters read? I would think the left one would read the same as the right one because the paths are the same length and the E field would be symmetrical between the sides. I don't see why the value of the resistance would affect the induced voltage since there is no current (switch open). The potential depends on the path length, not the value of the resistance.

If you mean "E field" in the bold line is the total E field, then I don't think the E field would be symmetrical. There are different kinds of E fields here. One is the induced E field. The other is the static E field due to charge redistribution at the boundary surface (When you connect a resistor to a wire, the resistor & the wire are 2 different mediums of 2 different conductivities). The static E field always vanishes in a circular integral, so that's why \oint E dl = \oint E_{induced} dl. But that doesn't mean E = E_{induced}.

The integral depends on the path for that reason, i.e. \oint E dl = \oint E_{induced} dl and induced E field is not consevative. The role of the resistance is that it makes the whole conducting medium inhomogeneous, and thus, leads to the presence of the static E field. With the presence of the static E field, the voltages read on two voltmeters are different, though the contribution to the voltage of the induced E field is the same.

I'm not sure about the timing of the experiment, as I watched it and joined this discussion a long time ago. The redistribution time of the charges for metal is around 10^-14s if I'm not wrong, but the resistors in the experiment are possibly made of a different material. Anyway, one thing for sure is the static E field does exist.

One theoretical way to check: the wire has nearly 0 resistance, so the total E field inside the wire must also be 0. The presence of another E field is thus essential to cancel out the induced E field.

Now suppose that the switch is closed and the experiment repeated. As soon as the magnetic field of the solenoid starts building up, an emf is induced in the circuit such that emf = \oint E\cdot dl = -d\phi/dt. Now the induced emf starts current flowing in the circuit. Suppose at time t very shortly after the solenoid current begins that total induced emf is 1 volt. The rate of change of that current is quite high at the beginning so there is a high back-emf (Lenz' law) opposing that increase in current so, while the induced voltage from the solenoid field at time t is 1 volt, the current is not 1/1000 A. = 1 mA. It is much less. The two voltages measured on the voltmeters cannot add up to 1 volt.

Then the self-inductance of the circuit must be taken into account. In this case, even if the rate of change of current inside the circuit is high, as the self-inductance of the circuit is very small, we cannot conclude anything about the total voltage built up inside the circuit. Of course, the 2 voltages measured cannot add up to 1V, but it can be 0.1234V as the way you anticipated, or 0.9999V as the way Lewin's advocates expect.

 

[yungman]

There is another thread here asking about back emf of an inductor. I wonder how Maxwell's equation treat this also.

The argument that copper wire is good conductor and field inside is almost zero and cannot have voltage across the wire cannot explain the voltage developed across an inductor without an induced voltage source. This again is another example that I don't understand. Also the voltage across the inductor is progressive also, so the PHYSICAL observation is the induced voltage is DISTRIBUTED along the whole path of the wire in an inductor.

Anyone have insight that can explain all these in electrodynamics point of view?

 

[hikaru1221]

There is another thread here asking about back emf of an inductor. I wonder how Maxwell's equation treat this also.

The argument that copper wire is good conductor and field inside is almost zero and cannot have voltage across the wire cannot explain the voltage developed across an inductor without an induced voltage source. This again is another example that I don't understand. Also the voltage across the inductor is progressive also, so the PHYSICAL observation is the induced voltage is DISTRIBUTED along the whole path of the wire in an inductor.

Anyone have insight that can explain all these in electrodynamics point of view?

The induced voltage is indeed contributed along the inductor, but that's NOT the only voltage here. Just as E field = induced E field + static E field, there is also voltage by the static E field.

P.S.: That argument comes straight from the equation J = \sigma E. How are you going to refute this equation?

 

[yungman]

The induced voltage is indeed contributed along the inductor, but that's NOT the only voltage here. Just as E field = induced E field + static E field, there is also voltage by the static E field.

P.S.: That argument comes straight from the equation J = \sigma E. How are you going to refute this equation?

The static E is small, we all know that, it is always there, nobody argue on this one. How about the induced voltage which is the big one.

 

[hikaru1221]

The static E is small

Please prove this. I never know that the static E field is really that small.
Many texts ignore this fact I believe.

 

[yungman]

Please prove this. I never know that the static E field is really that small.
Many texts ignore this fact I believe.

\vec E=\frac{\vec J}{\sigma} \approx \frac {\vec J}{5\times 10^7}

It is approximation only. I did not border to look up conductivity of copper.

I am referring to induced E when current change.

 

[hikaru1221]

The E in that equation is the total E field. You are not proving anything about the static E field in particular.

There is a model to prove that equation (E = J/sigma), and it starts with the force of the total E field (aside from the damping force) that exerts on electron. Google for Drude model, you will see it.

I also started with that equation (E = J / 10^7 or E = J / something very large) to prove that static E field cancels out induced E field. Static E field is never that small to be neglected in any case. It is the total E field that is small.

 

[yungman]

The E in that equation is the total E field. You are not proving anything about the static E field in particular.

There is a model to prove that equation (E = J/sigma), and it starts with the force of the total E field (aside from the damping force) that exerts on electron. Google for Drude model, you will see it.

I also started with that equation (E = J / 10^7 or E = J / something very large) to prove that static E field cancels out induced E field. Static E field is never that small to be neglected in any case. It is the total E field that is small.

I don't know exactly what static E you refer to, In this thread, static E is the longitudinal field of a wire developed by the voltage drop when current pass through. Other than that, I am really not interested in the detail definition.

My only interest here on this thread is how does electrodynamics accommodate the induced voltage. THis is the 700lb gorilla here. Maybe I should start a new thread as most people avoid this thread!

 

[hikaru1221]

static E is the longitudinal field of a wire developed by the voltage drop when current pass through.

No, it isn't. It's not about the definition either. It is due to charge cummulation. It happens all the time, even in the simplest circuit as a resistor connected to a DC power supply.

The way I see it, this static E field is the way to solve your gorilla issue. This is how the circuit and everything reacts to that induced voltage/ induced E field: it builds up static E field and, if inductance of the circuit is significant, its own B field and induced E field. The static E field is built in accordance with the intrinsic characteristics of everything involved (resistors, wire). The current is then built up in accordance with the total E field, by J = sigma*E.

 

[cabraham]

I don't know exactly what static E you refer to, In this thread, static E is the longitudinal field of a wire developed by the voltage drop when current pass through. Other than that, I am really not interested in the detail definition.

My only interest here on this thread is how does electrodynamics accommodate the induced voltage. THis is the 700lb gorilla here. Maybe I should start a new thread as most people avoid this thread!

Faraday: v = -N*d(phi)/dt.
Ohm: J = sigma*E.
Lorentz: F = q*(E + uXB).

The induced voltage is described in Faraday's Law, FL. But we must be careful. A time changing mag flux is related to the emf (voltage) induced in the loop per FL. But the flux "phi" is a net flux, not the external flux exciting the loop.

When the time varying, herein "ac", mag flux phi, excited the loop, there is charge motion per Lorentz force law, LFL. Free electrons in the wire are moved in a direction determined by E & B. E acts tangentially, B acts normally. Hence a rotational field condition is present & the electrons circulate around the loop. But hold on. The moving e- constitute a current, which produces another mag field. If the external mag flux density is called "Be", & internal is "Bi", then Bnet = B = Be + Bi. Of course, the law of Lenz, LL, tells us that the induced or internal B field opposes the external B in polarity.

In addition, we have another thing going on. As e- transit through the wire, they crash into lattice particles & lose some energy. This energy is radiated in the form of photon emission. It is around 5 to 7 micron in wavelength, & is felt as heat in the infrared region. Charges accumulate due to said collisions, & these charges have their own associated E field, since e- carry E fields due to their own charge.

This accumulation of e- charge gives rise to an ir-rotational (conservative) E field. This E field has no curl & is not an emf, but a drop. It is a polarizing type of force, its curl is zero, as it can not drive electrons around the closed loop.

So what is the voltage? How is it determined? The voltage V, is simply the line integral along a particular path of E*dl. But E has 3 components, the induced E field due to external B, the E field due to the current in the wire & its ac B field, Bi, & the static ir-rotational E field due to charge accumulation incurred via electron collisions w/ the lattice structure.

The voltage from a to b, is the line integral of the composite E fields along a chosen path. Inside the Cu wire, what is the voltage value, Vcu? We are measuring voltage from terminals a & b, w/ the path as the inside of the Cu winding.

The external ac mag flux density Be, gives rise to an induced E field, Ee, such that electrons in the Cu wire are moved along the wire, w/ said Be acting normal. This Ee is rotational.

But the induced current, Iloop, has its own B field, ac, we will call Bi. Bi is an ac field, & it has the opposite polarity of Be. It induces an E field & voltage or emf in the loop. This is the self-inductance of the loop. This is Ei, also rotational. For a very low loop resistance, the Bi cancels the Be almost entirely. What remains is the E field due to charges accumulating due to collisions between electrons & lattice structure. Call this one Ec. as it is due to charged particles.

Hence Enet = E = Ee + Ei + Ec. But Ee & Ei nearly cancel perfectly inside a low resistance conductor. If the loop were open, Ei tends toward zero, & the loop voltage is maximum due to no cancelling between Ee & Ei. Closing the loop via a resistive load results in current & a counter-balancing E & B fields, Ei & Bi. So inside the copper we still have Ec. If the entire loop was very low resistance so that Ee & Ei nearly cancel entirely, we still have Ec.

The Cu sec winding in my example is 0.1 ohm. The 10A current times the 0.1 ohm results in the voltage drop of 1.0V. The line integral of the composite E field along the Cu path results in Ee & Ei almost cancelling, & Ec*dl giving us around 1.0V. In other words, as soon as the ext mag field, Be, enters the Cu wire, current is induced. The induced current has a strong B field, Bi, that cancels the external, & equilibrium is reached.

You seem to be looking for the 120V induced emf inside the copper in distributed form. But don't forget that there is a counter-emf generated as well. The distributed emf sources are nearly perfectly canceled by the distributed counter-emf sources. But the Ec component does not get cancelled. It accounts for the 1.0V drop inside the Cu.

To better visualize this, consider a low resistance loop of wire, 0.010 ohm, closed & immersed in an ac mag field, Be. The Lorentz force moves the free electrons in the wire. This is current. But the current generates its own mag field, Bi. The law of Lenz tells us that they oppose each other. If the current is 1.0A, w/ a 0.010 ohm loop resistance, the voltage around the loop is 0.010 volt.

But we now open the loop, keeping the flux & area the same. The current plummets to near zero, but the voltage increases to 10V. Here, the ac mag flux produces an induced emf of 10V in the open state. When the loop closes, the net loop voltage is a mere 0.010V.

Why the difference? Of course, it is the cancellation. The external & internal parts of B & E account for the drastic difference in voltage between high & low impedance conditions. With a high-Z loop, the external B field is unopposed. Without induced current, the E field is due to the external B field, & the full voltage is realized since there is no loop current to cancel it.

When the loop resistance is low, the current generates cancellation of the external fields. The voltage is the line integral of all 3 phenomena. How can the Cu wire have just 1.0V, when the heater load has 119V, when their paths start & end at the same 2 points, a & b? They are in parallel, yet differing voltages are found.

Why? Answer is cancellation, & charge accumulation due to differing resistance values. I believe I've made my case, but someone other than me should affirm.

Claude

 

[yungman]

Thanks both of you, there is a lot of info here. I have to print them out, read first and think about all that. I have my own thing to study at the moment and can't put undivided attention to this right at the moment.

 

[yungman]

Faraday: v = -N*d(phi)/dt.
Ohm: J = sigma*E.
Lorentz: F = q*(E + uXB).

The induced voltage is described in Faraday's Law, FL. But we must be careful. A time changing mag flux is related to the emf (voltage) induced in the loop per FL. But the flux "phi" is a net flux, not the external flux exciting the loop.

When the time varying, herein "ac", mag flux phi, excited the loop, there is charge motion per Lorentz force law, LFL. Free electrons in the wire are moved in a direction determined by E & B. E acts tangentially, B acts normally. Hence a rotational field condition is present & the electrons circulate around the loop. But hold on. The moving e- constitute a current, which produces another mag field. If the external mag flux density is called "Be", & internal is "Bi", then Bnet = B = Be + Bi. Of course, the law of Lenz, LL, tells us that the induced or internal B field opposes the external B in polarity.

In addition, we have another thing going on. As e- transit through the wire, they crash into lattice particles & lose some energy. This energy is radiated in the form of photon emission. It is around 5 to 7 micron in wavelength, & is felt as heat in the infrared region. Charges accumulate due to said collisions, & these charges have their own associated E field, since e- carry E fields due to their own charge.

This accumulation of e- charge gives rise to an ir-rotational (conservative) E field. This E field has no curl & is not an emf, but a drop. It is a polarizing type of force, its curl is zero, as it can not drive electrons around the closed loop.
Don't see how charge accumulate, this is conductor, no charge can accumulate inside. Is all this conductance loss? Why is it irrotational? How do you determine which component of E is rotational and irrotational? Please give formula.
So what is the voltage? How is it determined? The voltage V, is simply the line integral along a particular path of E*dl. But E has 3 components, the induced E field due to external B, the E field due to the current in the wire & its ac B field, Bi, & the static ir-rotational E field due to charge accumulation incurred via electron collisions w/ the lattice structure.
Please give formulas of Ec, Ei and Ee respect to their source.
The voltage from a to b, is the line integral of the composite E fields along a chosen path. Inside the Cu wire, what is the voltage value, Vcu? We are measuring voltage from terminals a & b, w/ the path as the inside of the Cu winding.
Can you draw a diagram showing the path?
The external ac mag flux density Be, gives rise to an induced E field, Ee, such that electrons in the Cu wire are moved along the wire, w/ said Be acting normal. This Ee is rotational.

But the induced current, Iloop, has its own B field, ac, we will call Bi. Bi is an ac field, & it has the opposite polarity of Be. It induces an E field & voltage or emf in the loop. This is the self-inductance of the loop. This is Ei, also rotational. For a very low loop resistance, the Bi cancels the Be almost entirely. What remains is the E field due to charges accumulating due to collisions between electrons & lattice structure. Call this one Ec. as it is due to charged particles.

Hence Enet = E = Ee + Ei + Ec. But Ee & Ei nearly cancel perfectly inside a low resistance conductor. If the loop were open, Ei tends toward zero, & the loop voltage is maximum due to no cancelling between Ee & Ei. Closing the loop via a resistive load results in current & a counter-balancing E & B fields, Ei & Bi. So inside the copper we still have Ec. If the entire loop was very low resistance so that Ee & Ei nearly cancel entirely, we still have Ec.
Again please show in drawing for open and closed loop.
The Cu sec winding in my example is 0.1 ohm. The 10A current times the 0.1 ohm results in the voltage drop of 1.0V. The line integral of the composite E field along the Cu path results in Ee & Ei almost cancelling, & Ec*dl giving us around 1.0V. In other words, as soon as the ext mag field, Be, enters the Cu wire, current is induced. The induced current has a strong B field, Bi, that cancels the external, & equilibrium is reached.
I don't see this, Like all transformers, the secondary always has a range of voltage because Lens Law is a voltage equation. Output voltage varies with load but mainly due to internal resistance and also the the internal resistance of the primary. That if you draw more and more current from the secondary, you draw more and more current from primary and the effective primary voltage dropped due to loss in the resistance of the primary.
You seem to be looking for the 120V induced emf inside the copper in distributed form. But don't forget that there is a counter-emf generated as well. The distributed emf sources are nearly perfectly canceled by the distributed counter-emf sources. But the Ec component does not get cancelled. It accounts for the 1.0V drop inside the Cu.
I don't get this, why then you physically can measure 119V between the two terminal.
To better visualize this, consider a low resistance loop of wire, 0.010 ohm, closed & immersed in an ac mag field, Be. The Lorentz force moves the free electrons in the wire. This is current. But the current generates its own mag field, Bi. The law of Lenz tells us that they oppose each other. If the current is 1.0A, w/ a 0.010 ohm loop resistance, the voltage around the loop is 0.010 volt.
Again, in the drawing, show how you measure the 0.01V.
But we now open the loop, keeping the flux & area the same. The current plummets to near zero, but the voltage increases to 10V. Here, the ac mag flux produces an induced emf of 10V in the open state. When the loop closes, the net loop voltage is a mere 0.010V.
Where is the 10V coming from?
Why the difference? Of course, it is the cancellation. The external & internal parts of B & E account for the drastic difference in voltage between high & low impedance conditions. With a high-Z loop, the external B field is unopposed. Without induced current, the E field is due to the external B field, & the full voltage is realized since there is no loop current to cancel it.

When the loop resistance is low, the current generates cancellation of the external fields. The voltage is the line integral of all 3 phenomena. How can the Cu wire have just 1.0V, when the heater load has 119V, when their paths start & end at the same 2 points, a & b? They are in parallel, yet differing voltages are found.
Please show a drawing how you set up the probe to measure the 1V and 119V with two meters.
Why? Answer is cancellation, & charge accumulation due to differing resistance values. I believe I've made my case, but someone other than me should affirm.

Claude

I am loss after half way. I'll wait for more equation and drawing before I can continue. I still would like to know how you measure the 1V drop due to internal resistance in your transformer. I need to see the path how you measure the different voltages in the non conservative case you present.

I am making a first attempt to make sense of the E components:

\nabla \times \vec E_e =-\frac{\partial \vec B_e}{\partial t} \;\hbox { to get Ee from Be.}

\nabla \times \vec B_i = \mu_0\sigma_{Cu} \vec E_e + \mu_0\epsilon\frac{\partial \vec E_e}{\partial t} \;\hbox { to get Bi from Ee.}

\nabla \times \vec E_i =-\frac {\partial \vec B_i}{\partial t} \;\hbox { to get Ei from Bi}

 

[cabraham]

I am loss after half way. I'll wait for more equation and drawing before I can continue. I still would like to know how you measure the 1V drop due to internal resistance in your transformer. I need to see the path how you measure the different voltages in the non conservative case you present.

I am making a first attempt to make sense of the E components:

\nabla \times \vec E_e =-\frac{\partial \vec B_e}{\partial t} \;\hbox { to get Ee from Be.}

\nabla \times \vec B_i = \mu_0\sigma_{Cu} \vec E_e + \mu_0\epsilon\frac{\partial \vec E_e}{\partial t} \;\hbox { to get Bi from Ee.}

\nabla \times \vec E_i =-\frac {\partial \vec B_i}{\partial t} \;\hbox { to get Ei from Bi}

You're on the right track. I type slowly & long posts take a lot of time. I gave a xfmr example, but my explanation was eventually focused on a loop immersed in an ac B field, like that of an antenna receiving rf. A xfmr has one more thing going on.

A loop immersed in an rf B field in free space is subjected to induction. But said loop has an area which receives a specific amount of radiated power. This is induction w/ constant power. In the open circuit state, v = -N*d(phi)/dt. Also, phi = Ac*B, where Ac is the cross-sectional area of the loop, & B is mag flux density.

When the loop is open, Be, the external mag field, is related to the loop voltage Vloop, as follows. Vrms = Bpk/(4.443*f*Ac*N), where f is frequency, N is turns, per Faraday's law, FL. But if we close the loop in a high value of resistance R, we get current.

This current generates a field which opposes Be, so we call it Bi. I covered the rest previously. An equilibrium is reached when the loop resistance R is low enough so that the Bi cancels Be. Lowering R further does not increase the current. The voltage reduces as R is lowered, i.e. loop voltage decreases w/ decreasing R & increasing current.

It has to be this way per conservation of energy law, CEL. This is a constant power condition. The loop receives a limited amount of rf power, & the loop power cannot exceed the incident power per CEL. Hence Bi cancels Bi when R is low enough.

Now the xfmr is examined. When open secondary is measured, 120V appears at the terminals. Let's use these parameters for the xfmr including core. Vrms = 120V, Ac = (5cm X 5cm) 25 cm^2, f = 60 Hz, N = 120 turns both pri & sec, lc = core path length = 50 cm, Rsec = 0.1 ohm, & mu_r = relative permeability of core including incidental gap = 1000.

The B field in the core computes to 1.501 tesla per FL (15,010 gauss, a typical value for a grain oriented silicon steel material). To get H, we divide by mu, where mu = mu0*mu_r. Since NI = integral H*dlc, we get a magnetizing current of 0.498 amp, or 0.5A rounding off.

So we have a xfmr w/ sec open, 120V rms, & 0.5A magnetizing current, Imag. What happens when we connect the 11.9 ohm heater load across the secondary. The 0.1 ohm sec winding resistance is in series w/ the 11.9 ohm heater, for a total of 12.0 ohm, & the sec current, Isec = 10A. The terminal voltage drops by 1.0V to 119V rms.

The 1.5 tesla is the core when open is Be in this case. It requires an H to sustain it, w/ Imag of 0.5A. If the sec is loaded, that load current, induced by Be/Ee, tends to produce a mag flux, or "mmf", opposite in polarity to Be. This is Bi. Only 0.5A of counter-mmf will cancel the 1.5 tesla of Be. So where does the 120V come from, as you just asked?

A xfmr is not operating under a constant power condition like a loop in free space. A xfmr operates w/ constant voltage. The primary is connected across a good strong well-regulated constant voltage source, CVS. The power company goes through great effort to insure the voltage at our outlet is 120V rms.

As soon as load current is drawn at the sec, the counter-mmf produces Bi cancelling Be, resulting in a drop in terminal voltage. But the xfme primary is connected to a CVS. Said CVS then outputs an increase in current which counters the counter-mmf. The additional primary current provides "counter-counter-mmf". Just as the counter-mmf (or "Bi" if you prefer) resulted in counter-emf & a drop in voltage, the counter-counter-mmf produces a counter-counter-emf & an increase in voltage.

As long as the primary is excited by a good solid CVS which has the power capability to meet the load demands, said CVS will offer any current necessary to keep Vpri at 120V rms. Thus the cancellation of Be by Bi, is countered by increased Ipri.

But the mag flux still cannot enter the Cu sec winding to any large degree. Since the sec Cu resistance & the heater load are in series, their current is identical. Hence the 120V is divided between the 0.1 ohm & 11.9 ohm. When current exists in 2 different resistances in series, the higher resistance material incurs more electron to latiice collisions, & more accumulated charges. The charges provide their E field, Ec. When all 4 components of E are evaluated, we get 1.0V in the Cu sec, & 119V in the heater load.

In a nutshell, the CVS sets up a core flux, which sets up a sec E field & voltage. When loaded, the sec current produces Bi/Ei which cancels the Be/Ee. The CVS then forces an equilibrium condition by providing just enough primary current so thet Vloop = 120V. It's a CVS, that is what it does. Lattice collisions take place more frequently in the higher resistance medium. But being in series the current in each of the 2 media must be equal. Hence charges build up & the E field due to charges, Ec, adds or subtracts to/from Ee, Ei, & counter-Ee.

Dr. Lewin has a good paper on Ec which I'll dig up & post. Did this help?

Claude

 

[yungman]

Thanks for your time. Again, I'm going to have to take time to read before I reply. Can you give some drawings if you have time at the mean time.

One thing concern me so far in this analysis is: If I follow this theory, then Ei and will cause another Bi2 and then Bi2 will cause Ei2 and Ei2 will cause Bi3 and to Ei3 and so on. So it is going to be an approximation of an infinite series.

I am not saying I agree with you, You are knowledgeable, I want to try to understand your point of view before going any further. Actually I am going to stop my own study and review the Lenz law today first because the EM books don't really go too deep into Lenz. I am going to read through the ED book today.

Thanks

 

[cabraham]

Thanks for your time. Again, I'm going to have to take time to read before I reply. Can you give some drawings if you have time at the mean time.

One thing concern me so far in this analysis is: If I follow this theory, then Ei and will cause another Bi2 and then Bi2 will cause Ei2 and Ei2 will cause Bi3 and to Ei3 and so on. So it is going to be an approximation of an infinite series.

I am not saying I agree with you, You are knowledgeable, I want to try to understand your point of view before going any further. Actually I am going to stop my own study and review the Lenz law today first because the EM books don't really go too deep into Lenz. I am going to read through the ED book today.

Thanks

I see your point, & I feel you are thinking in good terms, but the underlined "cause" words in the highlighted text give me concern. I believe, & most of the science community believes, that E & B are mutually inclusive, cannot exist independently, & that neither can be the cause nor the effect of the other. Special relativity describes E & B forces both as Coulomb interaction forces between charges taking special relativity into account. E & B are 2 views of the same action from different frames of reference.

But you are correct that the net E field is the sum of Ee from the distant source, Ei from the induced current, counter-Ee from the CVS exciting the primary, & Ec due to charge build-up in boundary regions. The Ec concept is explained by none other than Prof. Lewin in the attached lecture paper. It should help immensely. BR.

Claude

 

[yungman]

I understand about the chicken and egg thing. I was just following your notion of Be cause Ee and then Bi and Ei step by step. Which make sense.

Too much reading materials! They really don't go deep into Lenz in EM books, they more dive into EM waves and Tx lines. I am reading the Lenz in Griffiths and talking about 3 E also. I might not be back tonight, too much reading!

One thing I am thinking.

Emf = -\frac {d\Phi}{d t}

This show flux induce a voltage, not current like what you use. You seems to start with say the secondary providing 10A with internal resistance of 0.1 ohm. then you show the short circuit around the secondary is 1V where it is 10A X 0.1 ohm!


Thanks

 

[cabraham]

I understand about the chicken and egg thing. I was just following your notion of Be cause Ee and then Bi and Ei step by step. Which make sense.

Too much reading materials! They really don't go deep into Lenz in EM books, they more dive into EM waves and Tx lines. I am reading the Lenz in Griffiths and talking about 3 E also. I might not be back tonight, too much reading!

One thing I am thinking.

Emf = -\frac {d\Phi}{d t}

This show flux induce a voltage, not current like what you use. You seems to start with say the secondary providing 10A with internal resistance of 0.1 ohm. then you show the short circuit around the secondary is 1V where it is 10A X 0.1 ohm!


Thanks

Well. if the loop resistance is non-infinite & non-zero, then current & voltage are both induced. For a shorted loop we have I = integral H*dl. For an open loop emf = -d(phi/dt).

If the loop resistance is relatively high then phi_e is all that accounts for the emf. But w/ a low value of loop resistance, phi_i cancels phi_e. Phi = Ac*B. The Lorentz force states that there is a force acting on electrons, moving them. In the process of moving they incur collisions & a voltage drop occurs. Also, while moving they generate an internal B field, Bi which generates another emf.

So the emf equation you cited has multiple parts. Of course emf = -d(phi)/dt, but remember that phi is the external, plus internal, plus another external if the problem involves a CVS as a source, plus the E field due to charges Ec.

The emf equation is correct but there is a lot going on here.

Claude

 

[yungman]

In the shorted circuit, the resistance of the primary come into play, you draw so much current the primary voltage drop due to the drop in the primary resistance. This together with the secondary resistance will limit the max current. But that does not mean that it become a current equation. If you put very heavy primary winding and secondary, I don't think you can use the idea of set current of say 10A and get only 0.01 V in the secondary.

Can you give the diagram how you measure the different voltage drop with different path. that would be the ultimate proof. In order to see that you get 119V in one path and 1V in the second, we are going to have to be able to read that physically.

 

[yungman]

Hey Claude

Thanks for your time. I understand a lot more about Faradays law in the last two days from your posts and from the book I read. I started another post concentrate on what you wrote because I think it warrant more discussion that is not related to this professors case.

I understand the constant flux \Phi in antenna vs the transformer plugging into the wall plug. Please join in the other post to continue that point. And I think what you said has no bearing to Levin's case.



Now back to the MIT professor.

I think the transformer is a lot more easier to talk about than the Levin's single loop case. The reason is because both example reflect the same phenomena. In Levin case, he adjusted the input flux to get 1V total output so it's like your transformer that the power company maintain constant primary voltage and increase current to give constant voltage at the secondary and maintain 1V.

The difference is in your case, your secondary is say 120 turns, so the induced voltage onto the measuring probe's ground lead only is 1/120 of the overall voltage. So the measurement has only about 1% error vs the single loop secondary of Levin's experiment. The ground lead of the probe would produce 100% error like what I shown in #224 and #227.


Now this is my observation: I don't see where in your transform example give any voltage source to give 120V. In order to draw out the circuit diagram, you are going to have to give the source of the voltage. All the explanation of the magnetic field and electric field are like creating voltage source and counter voltage sources and so on. But how do you draw it in the circuit diagram? Just like Antiphon said, Levin mix circuit drawing and real EM circuit and it does not work that way. If you have a voltage, you have to represent it in the circuit diagram. Levin is wrong to take the circuit diagram literaly and continue on with the physical observation.

In order for Levin or anyone else to justify the path dependent, they have to justify where the voltage come from. It is ABSOLUTELY not enough to just say we "STIPULATE" the voltage exist and then turn around and use the circuit diagram without it and justify the path dependent thing. You want to justify the non conservative thing, you have to justify the voltage source ( or the lack of it). If you want to draw the circuit diagram to represent the real physical circuit, you are going to have to draw where the voltage come from.

Also, the good thing about using the transformer, we have absolutely no problem doing any measurement now. Now let's redraw the circuit, redefine the point A and D and then let's do the measurement of the two path, they WILL be the same using the transformer in the middle. There is no if and buts about this, you cannot treat the transformer as a single note D anymore. In Levin's case, he need to draw in the transformer before he can use the circuit diagram to justify the path dependent statement. This is what I have been arguing about for a long long time, maybe not in these exact words...Because never in my wildest imagination that ED people here don't consider induced voltage source and we end up going around and around, I kept drawing circuit based of equivalent voltage sources and a lot of you guys just ignor my drawing!




Well? Anyone else, I know from looking and the number of people reading this thread that there are plenty of people interested in this. What is your opinion?

 

[yungman]

This is the drawing of Levin's circuit and Cabraham transformer. It is basically the same circuit.

I did not draw the primary of the transformer, it is understood the primary is just to provide the \Phi and like the power company, Levin adjust the driver of the solenoid to get 1V on the two resistors. This constant output voltage really simplify all the arguments of Faraday's law, secondary internal B and E because both case adjust the primary to get the desired voltage. The theory become very very simple...It become a VOLTAGE SOURCE!

As you can see in part A, the circuit that Levin drew should really be the right hand side drawing that include a transformer loop that have flux going through it and voltage induced on the loop.


In B, this is Cabraham's transformer circuit that he described. The internal resistance of the secondary is 0.1 ohm and the load resistance is 11.9 ohm. The circuit draw 10A and the voltage across the load is 119V. You can see the left side of B is drawn without any voltage source and treat the secondary of the transformer as a note D just as Levin drawn in his circuit diagram. But if you draw the transformer in like on the right side, then the circuit is completed.


Bottom line, Levin is WRONG to use this as an example to prove non conservative voltage that is path dependent because he left out the transformer part in the circuit. And further he call the transformer as note D.

Feel free to comment on my finding.




To StevenB:

With the 120 turns transformer secondary. There should be no problem measuring the voltage using a probe no matter how it swing, what direction the probe come in. BECAUSE, the loop formed by the probe is only a single turn only, so it would introduce about 1% of errour however which way you swing. AND you know and I know that now, if you tape on anywhere of the secondary of the transformer, you will get different voltage in the measurement. It would not be like the experiment we did that we are trying to measure along the single turn transformer while the loop formed by the probe ground keep fighting us.

I did not work on finding a way to measure the experiment because I figure no matter how I do it, there is always a reason to argue against and it's not fruitful. I am thankful someone dug this up and ANTIPHON shine the light on the difference between circuit diagram and real components.

 

[cabraham]

I'll draw a pic later. I've been tidying up the house today because I have guests arriving tonight. Been quite busy. Tonight or tomorrow, I'll have an illustrative pic. But here is a quick comment.

The pic you showed has the 0.1 ohm sec Cu resistance & the xfmr sec inductor depicted as separate lumped parameters. You earlier argued that the reality is distributed parameters including emf sources.

The emf in the xfmr case is forced to be roughly the value of CVS exciting the primary. Ee & Ei will oppose each other, but the CVS provides whatever primary current is needed to force a flux density of 1.5 tesla, resulting in 120V rms terminal voltage. But the voltage along the Cu path is 1.0V, & along the heater path is 119V. Why?

I'll have a pic later. For now please refer to the Dr. Lewin paper I recently posted. At the Cu-heater boundary, we have differing resistances. But the current in the Cu & the heater are the same, being connected in series.

So per Ohm's Law, OL. Jcu = sigma_cu*Ecu, & Jhtr + sigma_htr*Ehtr. But we know that Jcu = Jhtr = J. So that sigma_cu*Ecu = J = sigma_htr*Ehtr. We know that the sigma values differ, as Cu has much lower resistivity. But the CVS driving the xfmr primary forces an emf of 120V rms. We know this. This emf is distributed, i.e. 1.0 volt/turn. But why is there 1.0V across the Cu inside the wire, & 119V across the heater?

What if the sec winding was made with high resistance wire, so that both the Cu sec & the heater were 6.0 ohms each? There would be no difference in terminal voltage regardless of path. It would be 60V across the Cu & 60V across the heater. The 120V emf, forced by the CVS at the xfmr primary is distributed around the loop, & since the resistances are identical for each path, no Ec exists, i.e. the E field in each section is due to emf only, no static charges are present.

But if sec Cu is 0.10 ohm, heater is 11.9 ohm, we still have 120V emf for the sec loop regardless of resistance values. But due to the differing resistances, there is a charge build-up at each interface. Refer to Dr. Lewin's diagram. One boundary (between Cu & heater) has an accumulation of "+ve" charge, the other boundary "-ve" charge.

These discrete charges have their own E field, Ec. The Ec has a polarity as follows. In the Cu, Ec is oriented so as to subtract from Ee. In the heater Ec adds to Ee. So the loop emf stays at 120V because this Ec field has zero curl. Going around the loop integrating Ec only results in a rise & a drop of equal magnitude. Around the loop, Ee integrates to 120V, & Ec to zero.

In the Cu, Ee integrates to half (assuming symmetry) which is 60V, & Ec to -59V. The net voltage in the Cu is the difference of 1.0V. But in the heater, Ee integrates again to half, or 60V, but Ec integrates to +59V. The net voltage in the heater is the sum of 119V.

Meanwhile, the CVS is oblivious to Ec. Take the 2 cases I detailed, 6.0 ohm Cu sec with 6.0 ohm heater, & the other case being 0.1/11.9 ohm resp. The net load is 12 ohm, 10A, either way, likewise for the 120V emf. The CVS & the process of induction are oblivious to the exact resistance distribution across the secondary. In both cases, the emf is 120V, & the CVS, the core flux, etc. is unaware of how the 120V is distributed between the Cu resistance & load resistance.

Nature fixes this with the accumulation of static charges at the interface between the 2 media. This charge field has its own Ec field. The Ee type of field has curl. Around a loop it has a non-zero integral & non-zero emf. But the Ec field is non-curly, having zero net emf around the loop. In other words the CVS is oblivious to Ec. Ec has zero curl & hence it cannot "mess up" the 120V emf, mandated by laws of induction & the nature of a CVS.

But the Ec field exactly fixes the voltage distribution while preserving Ohm's Law, Faraday, Ampere, conservation of energy, Lenz, etc. By definition, the voltage from xfmr sec terminals a to b, is the integral of the composite E field, whose parts are Ee, Ei, counter-counter Ee provided by the CVS, as well as the all-important Ec.

Do NOT neglect Ec. It does not show up in the induction laws, & has no interaction w/ the CVS. It merely distributes the 120V emf in accordance w/ the resistances preserving Ohm's law.

J = sigma*E, so that if J is the same for 2 regions, which it is, & E were the same, then we would have a paradox. How can J & E both be the same while the sigma values differ for each region. The answer is that if we combine Ee & Ei & the CVS Ee all into a composite Ee', that value of Ee' is uniform. It would integrate to the same value in Cu & heater. Hence we get a paradox. The energy per unit charge to transport charge through copper is but 1.0V, vs. 119V for heater.

Ec fixes all that. Symmetry would result in 60V in each section. Ec provides -59V in the Cu & +59V in the heater. All laws are upheld.

Not so intuitive is it? Great minds stumble with this concept. It takes a lot of thought to sort it out. Prof. Lewin is spot on dead right!. Cheers.

Claude

 

[stevenb]

Not so intuitive is it? Great minds stumble with this concept. It takes a lot of thought to sort it out. Prof. Lewin is spot on dead right!.

I second that comment.

I'll add one of my own too. Thanks to the great mind of Faraday, we don't need a great mind, nor should we stumble, when we study this subject. We need only trust in his Law. And, even if we want to get into these nitty gritty details, how much easier even that will be if we start with an acceptance that Faraday's Law trumps our intuition.

 

[cabraham]

I second that comment.

I'll add one of my own too. Thanks to the great mind of Faraday, we don't need a great mind, nor should we stumble, when we study this subject. We need only trust in his Law. And, even if we want to get into these nitty gritty details, how much easier even that will be if we start with an acceptance that Faraday's Law trumps our intuition.

Faraday was inded a great mind. Lest we forget, he had a high school education, the son of a blacksmith. He read books on his own, employed in a book store. Books were costly & hard to obtain in those days.

Faraday's & all other verified laws trump our intuition 24/7/365, i.e Ohm, Ampere, energy, Kirchoff current/voltage, etc. Relying on intuition is the most dangerous think I know of. The problem w/ intuition is that it seems so logical to us. But we only factor in the truths we are aware of, omitting details we are unaware of, but are pertinent nonetheless. Regarding this Prof. Lewin problem, the fact that static charges accumulate near the boundary region of the 2 media, & form another E field, Ec, that is conservative, & adds/subtracts w/ the Ee field, is well --- who'd have thunk it! But the presence of said E field dispels any paradox & upholds OL, CEL, FL, LL, LFL, AL, etc.

Science is amazing!

Claude

 

[yungman]

I'll read yours and Levin's tonight. I have to admit, I am a lot more open mind reading your's or any reference materials or textbooks. I just find it hard to read anything from Levin anymore. But I'll read it with as open a mind as possible.

BTW, I pull out your two posts and started another thread. Please comment on that one, I just found that very interesting and start thinking more, I am not stating any opinion, just want to listen to people.

 

[yungman]

I see your point, & I feel you are thinking in good terms, but the underlined "cause" words in the highlighted text give me concern. I believe, & most of the science community believes, that E & B are mutually inclusive, cannot exist independently, & that neither can be the cause nor the effect of the other. Special relativity describes E & B forces both as Coulomb interaction forces between charges taking special relativity into account. E & B are 2 views of the same action from different frames of reference.

But you are correct that the net E field is the sum of Ee from the distant source, Ei from the induced current, counter-Ee from the CVS exciting the primary, & Ec due to charge build-up in boundary regions. The Ec concept is explained by none other than Prof. Lewin in the attached lecture paper. It should help immensely. BR.

Claude

I read the lecture notes. So what Levin said is the E only concentrate in the part of the loop that is of highest resistance and the rest of the loop that make up of wires ( very low resistance) has very little electric field. That he use the example of the two half of the loop make of R1 and R2 where R2>R1 and show less voltage drop (itex] V=\int E \cdot d \vec l [/itex] across R1 ( left side ) than right side of the loop.

I understand all the formulas he gave, but when come the part of the charges build up at the junction, I don't follow at all and there is no formulas to support this.


So basically what Levin claimed is in an open circuits, there is no voltage across the two terminals ( A and B) on the secondary of the transformer. Voltage only created when you put a resistor across the terminals. So there is no equivalent voltage source in the transformer or matter of fact in Levin's experiment.

Has this gone through peer review? Any other reputable people support this? Can you sight some other articles to support this or IEEE papers? Or in AIP papers?

 

[yungman]

I'll draw a pic later. I've been tidying up the house today because I have guests arriving tonight. Been quite busy. Tonight or tomorrow, I'll have an illustrative pic. But here is a quick comment.

The pic you showed has the 0.1 ohm sec Cu resistance & the xfmr sec inductor depicted as separate lumped parameters. You earlier argued that the reality is distributed parameters including emf sources.

The emf in the xfmr case is forced to be roughly the value of CVS exciting the primary. Ee & Ei will oppose each other, but the CVS provides whatever primary current is needed to force a flux density of 1.5 tesla, resulting in 120V rms terminal voltage. But the voltage along the Cu path is 1.0V, & along the heater path is 119V. Why?

I'll have a pic later. For now please refer to the Dr. Lewin paper I recently posted. At the Cu-heater boundary, we have differing resistances. But the current in the Cu & the heater are the same, being connected in series.

So per Ohm's Law, OL. Jcu = sigma_cu*Ecu, & Jhtr + sigma_htr*Ehtr. But we know that Jcu = Jhtr = J. So that sigma_cu*Ecu = J = sigma_htr*Ehtr. We know that the sigma values differ, as Cu has much lower resistivity. But the CVS driving the xfmr primary forces an emf of 120V rms. We know this. This emf is distributed, i.e. 1.0 volt/turn. But why is there 1.0V across the Cu inside the wire, & 119V across the heater?

What if the sec winding was made with high resistance wire, so that both the Cu sec & the heater were 6.0 ohms each? There would be no difference in terminal voltage regardless of path. It would be 60V across the Cu & 60V across the heater. The 120V emf, forced by the CVS at the xfmr primary is distributed around the loop, & since the resistances are identical for each path, no Ec exists, i.e. the E field in each section is due to emf only, no static charges are present.

But if sec Cu is 0.10 ohm, heater is 11.9 ohm, we still have 120V emf for the sec loop regardless of resistance values. But due to the differing resistances, there is a charge build-up at each interface. Refer to Dr. Lewin's diagram. One boundary (between Cu & heater) has an accumulation of "+ve" charge, the other boundary "-ve" charge.

These discrete charges have their own E field, Ec. The Ec has a polarity as follows. In the Cu, Ec is oriented so as to subtract from Ee. In the heater Ec adds to Ee. So the loop emf stays at 120V because this Ec field has zero curl. Going around the loop integrating Ec only results in a rise & a drop of equal magnitude. Around the loop, Ee integrates to 120V, & Ec to zero.

In the Cu, Ee integrates to half (assuming symmetry) which is 60V, & Ec to -59V. The net voltage in the Cu is the difference of 1.0V. But in the heater, Ee integrates again to half, or 60V, but Ec integrates to +59V. The net voltage in the heater is the sum of 119V.

Meanwhile, the CVS is oblivious to Ec. Take the 2 cases I detailed, 6.0 ohm Cu sec with 6.0 ohm heater, & the other case being 0.1/11.9 ohm resp. The net load is 12 ohm, 10A, either way, likewise for the 120V emf. The CVS & the process of induction are oblivious to the exact resistance distribution across the secondary. In both cases, the emf is 120V, & the CVS, the core flux, etc. is unaware of how the 120V is distributed between the Cu resistance & load resistance.

Nature fixes this with the accumulation of static charges at the interface between the 2 media. This charge field has its own Ec field. The Ee type of field has curl. Around a loop it has a non-zero integral & non-zero emf. But the Ec field is non-curly, having zero net emf around the loop. In other words the CVS is oblivious to Ec. Ec has zero curl & hence it cannot "mess up" the 120V emf, mandated by laws of induction & the nature of a CVS.

But the Ec field exactly fixes the voltage distribution while preserving Ohm's Law, Faraday, Ampere, conservation of energy, Lenz, etc. By definition, the voltage from xfmr sec terminals a to b, is the integral of the composite E field, whose parts are Ee, Ei, counter-counter Ee provided by the CVS, as well as the all-important Ec.

Do NOT neglect Ec. It does not show up in the induction laws, & has no interaction w/ the CVS. It merely distributes the 120V emf in accordance w/ the resistances preserving Ohm's law.

J = sigma*E, so that if J is the same for 2 regions, which it is, & E were the same, then we would have a paradox. How can J & E both be the same while the sigma values differ for each region. The answer is that if we combine Ee & Ei & the CVS Ee all into a composite Ee', that value of Ee' is uniform. It would integrate to the same value in Cu & heater. Hence we get a paradox. The energy per unit charge to transport charge through copper is but 1.0V, vs. 119V for heater.

Ec fixes all that. Symmetry would result in 60V in each section. Ec provides -59V in the Cu & +59V in the heater. All laws are upheld.

Not so intuitive is it? Great minds stumble with this concept. It takes a lot of thought to sort it out. Prof. Lewin is spot on dead right!. Cheers.

Claude

You are basically repeating Levin's notes. So you claim that the reason you can measure 119V across the terminal is only because of the charge build up at the junction that make up 118V and the whole secondary only drop 1V? I don't get the charge buildup at the junction at all. Is there any experiment proof on this? Or is this his theory? Is there others with authority in this field support this?


This is pretty much the bottom line of the arguement. I sure have never seen anything like this in books and I don't see any proof or formulas to support this that the charge make up the voltage. How come nobody else want to join in the discussion. I am sure most of your guys has better theory background than me!

 

[hikaru1221]

Yungman, don't refuse the fact that the argument comes straight from J = E*sigma. And that formula certainly is somewhere in Cheng's or Griffiths'. That's the theoretical base. Most books don't touch on this kind of phenomenon I believe, but that doesn't mean it doesn't exist.

This IEEE paper's abstract, though barely related to this discussion, points out that "Different surface treatments have further been performed in order to study the influence of space charge accumulation in the boundary layer", and evidences the existence of the charge accumulation at the boundary of 2 different media.
http://ieeexplore.ieee.org/xpl/freeabs_all.jsp?arnumber=522993
I know you may say it exists but is small & negligible. You can check the theory yourself, or stay happy with your own theory.

I also did provide you a document & an experiment video from some EE professors of MIT in support of Lewin's explanation a long time ago, but you probably didn't bother looking at it. This is the document & the video:
http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-013-electromagnetics-and-applications-fall-2005/textbook-with-video-demonstrations/
The download link of the document is at "PDF" at chapter 10 when you scroll down - see example 1 from page 2 to 4 of the document. Below that, at section 10.0.1, there is a downloadable video, which demonstrates the experiment of the example 1 of the document. And I don't think that these professors don't know what IEEE is.
(that doesn't mean I favor MIT. MIT OCW is the only place where I can find the supporting document)

So basically what Levin claimed is in an open circuits, there is no voltage across the two terminals ( A and B) on the secondary of the transformer. Voltage only created when you put a resistor across the terminals. So there is no equivalent voltage source in the transformer or matter of fact in Levin's experiment.

This is not true. Lewin didn't conclude that, and even the theory won't end up at such conclusion. Ec (static E field) can be built up everywhere, even in the air, the same as Ei (induced E field). For an open circuit, we can regard the gap as a resistor of very high resistance. Extrapolating from the conclusion of Lewin, which is that total voltage across the resistor of higher resistance is higher, we can see that this air resistor takes all the voltage.

Voltage is created with total E field. The induced E field Ei is independent of the resistor. The static E field Ec is dependent on Ei and the resistor. Without resistor and thus Ec, there is still voltage by Ei. With resistor and thus Ec, there is also voltage, but with different distribution.

P.S.: To ellaborate on the point of J = E*sigma, that's not the only thing to apply. You will need Gauss law. Applying J = E*sigma, you arrive at that total E at 2 different media are different, if J is the same. Applying Gauss law, you arrive at that at the boundary surface, there is charge accumulation.

 

[yungman]

Yungman, don't refuse the fact that the argument comes straight from J = E*sigma. And that formula certainly is somewhere in Cheng's or Griffiths'. That's the theoretical base. Most books don't touch on this kind of phenomenon I believe, but that doesn't mean it doesn't exist.

This IEEE paper's abstract, though barely related to this discussion, points out that "Different surface treatments have further been performed in order to study the influence of space charge accumulation in the boundary layer", and evidences the existence of the charge accumulation at the boundary of 2 different media.
http://ieeexplore.ieee.org/xpl/freeabs_all.jsp?arnumber=522993
I know you may say it exists but is small & negligible. You can check the theory yourself, or stay happy with your own theory.

I also did provide you a document & an experiment video from some EE professors of MIT in support of Lewin's explanation a long time ago, but you probably didn't bother looking at it. This is the document & the video:
http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-013-electromagnetics-and-applications-fall-2005/textbook-with-video-demonstrations/
The download link of the document is at "PDF" at chapter 10 when you scroll down - see example 1 from page 2 to 4 of the document. Below that, at section 10.0.1, there is a downloadable video, which demonstrates the experiment of the example 1 of the document. And I don't think that these professors don't know what IEEE is.
(that doesn't mean I favor MIT. MIT OCW is the only place where I can find the supporting document)



This is not true. Lewin didn't conclude that, and even the theory won't end up at such conclusion. Ec (static E field) can be built up everywhere, even in the air, the same as Ei (induced E field). For an open circuit, we can regard the gap as a resistor of very high resistance. Extrapolating from the conclusion of Lewin, which is that total voltage across the resistor of higher resistance is higher, we can see that this air resistor takes all the voltage.

Voltage is created with total E field. The induced E field Ei is independent of the resistor. The static E field Ec is dependent on Ei and the resistor. Without resistor and thus Ec, there is still voltage by Ei. With resistor and thus Ec, there is also voltage, but with different distribution.

P.S.: To ellaborate on the point of J = E*sigma, that's not the only thing to apply. You will need Gauss law. Applying J = E*sigma, you arrive at that total E at 2 different media are different, if J is the same. Applying Gauss law, you arrive at that at the boundary surface, there is charge accumulation.

Thanks

I have to take a look at chapter 10 first.

Alan